Problem: An object is attached by a string to the end of a spring. Fang throws the object upwards and starts a stopwatch at $t=0$ seconds. The object starts oscillating vertically in a periodic way that can be modeled by a trigonometric function. The object's average height is $-20\text{ cm}$ (measured from the top of the spring). It first achieves that average height on the way up at $t=0.2$ seconds, and then again every $2$ seconds. The object's maximum and minimum heights are each $5\text{ cm}$ from its average height. Find the formula of the trigonometric function that models the height $H$ of the weight $t$ seconds after Fang started the stopwatch. Define the function using radians. $ H(t) = $ What is the height of the object after $0.6$ seconds? Round your answer, if necessary, to two decimal places. $ $
Let's start by writing a formula for the distance $u$ seconds after it first reaches its average height ( $u$ seconds after $t = 0.2$ ). Both sine and cosine can be used to model periodic contexts. We can decide which is better fitting by considering the $y$ -intercept. The sine function intercepts the $y$ -axis at its midline, and the cosine function intercepts the $y$ -axis at its peak. That way, we know the object's height is increasing past its average value when $u = 0$, so we can use a sine function, which is also increasing past its average value when $u = 0$. The amplitude of the height function is the distance between its average and maximum values, or $5\text{ cm}$. The period is $4$ seconds, since a sine function reaches its average value twice in every period (at $0$ and $\pi$ ). The midline is the line at the average value, or $y = -20\text{ cm}$. Since the ordinary sine function $f(u) = \sin u$ has period $2\pi$, midline $y = 0$, and amplitude $1$, we need to stretch it horizontally by a factor of ${\dfrac{4}{2\pi}}$, stretch it vertically by a factor of ${5}$, and move it down ${20}$ units: $ H(u) = {5}\sin\left({\dfrac{2\pi}{4}}u\right) - {20}$ Since the object passes its midline $0.2$ seconds after the stopwatch is started, $t$ seconds after the stopwatch is started is $t - 0.2$ seconds after it passes its midline. So $u = t - 0.2$ : $ H(t) = {5}\sin\left({\dfrac{2\pi}{4}}(t- 0.2)\right) - {20}$ When $t = 0.6$, the height of the object is $ \begin{aligned}H(0.6) &= {5}\sin\left({\dfrac{2\pi}{4}}(0.6- 0.2)\right) - {20} \\ &\approx -17.06\end{aligned}$ A correct formula for $H(t)$ is: $ H(t) = 5\sin\left(\dfrac{2\pi}{4}(t- 0.2)\right) - 20$ The height of the object after $0.6$ seconds is: $ -17.06\text{ cm}$ This is the graph of the function, with the object's height after $0.6$ seconds marked.